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Tangent plane to a level surface

mark

Find an equation of the plane tangent to the surface

x^2 + z^3 = z\,y

at the point (0,1,1).

mearing

I started by setting the surface in terms of x and z:

y(x,z)=\frac{x^2}{z}+z^2

Then found the partial derivatives:

f_{x}=\frac{2x}{z}

f_{z}=x^2+2z

f_{x}(0,1,1)=0 \space\space f_{z}(0,1,1)=2

The formula for the plane:

L(x,z)=1+2(z-1)

mark

@mearing I like this! BUT, I don’t think you can plan on being able to solve for one variable in terms of the others on the quiz. I added a problem here to illustrate.

dspivey

I got the equation (y-1)+(z-1)=0
From the partials in respect to x, y and z.

fcarrill

here is my idea of solving this
x^2+z^3=zy
=>x^2+z^3-zy=0 subtract zy from both sides
\nabla f<2x,-z,3z^2-y> got the gradiant for x,y,z plugged in 0,1,1
=<0,-1,2>
so for my equation i got 0(x-0)-1(y-1)+2(z-1)=0